For this case we have the following system of equations:
[tex]y = x-2\\y = x ^ 2-5x + 6[/tex]
Matching we have:
[tex]x ^ 2-5x + 6 = x-2[/tex]
We manipulate algebraically:
[tex]x ^ 2-5x-x + 6 + 2 = 0\\x ^ 2-6x + 8 = 0[/tex]
To factor, we look for two numbers that, when multiplied, result in +8 and when added, result in -6. These numbers are -4 and -2.
[tex](x-4) (x-2) = 0[/tex]
Thus, we have to:
[tex]x_ {1} = 4\\x_ {2} = 2[/tex]
We find the values of the variable y:
[tex]y_ {1} = x_ {1} -2 = 4-2 = 2\\y_ {2} = x_ {2} -2 = 2-2 = 0[/tex]
So the solutions are:
(4,2) and (2,0)
Answer:
Option D
See attached image
